Indeed, if x is an element of Q, then |f(x)| = |x^2| &lt; |x| &lt; epsilon, since |x|&lt;1.

if x is an element of R-Q, then |f(x)| = 0 &lt; epsilon.

Thus f is continuous at c=0.

2) f is also differentiable at c=0, in the sense that

lim_{x-&gt;0} ( f(x)-f(0) ) / x = 0.

Indeed, fix

epsilon&gt;0and put delta = epsilon.

Then for any x with

|x|&lt;deltawe have that | (f(x)-f(0)) / x | &lt; epsilon.

Indeed, if x belongs to Q-0, then

| [ f(x)-f(0) ] / x | = | (x^2 - 0) /x | = |x| &lt; epsilon

and if x belongs to R-Q, then f(x)=0, and

| (f(x)-f(0)) / x | = | (0 - 0)/x | = 0 &lt; epsilon

Thus lim_{x-&gt;0} [ f(x)-f(0) ] / x = 0.

3) On the other hand f is not continuous and not differentiable at any other point c&lt;&gt;0.Indeed, if c&lt;&gt;0, then take two sequences (a_i) and (b_i) converging to c such that each a_i is raitonal, i.e. belogns to Qand each b_i is irrational, i.e. belogns to R-Q.